How can a $C^1$-continuous surface have infinite curvature?

Short version:

Apparently it is possible for a $C^1$-continuous parametric surface to (locally) have infinite (Gaussian) curvature. I find this quite counter-intuitive, because I always thought that only a $C^0$-continuous surface (i.e. containing a kink) can have infinite curvature. What am I missing?

Long version:

Since the curvature of a curve is defined as $1/r$, where $r$ is the radius of the circle that best describes the curve at the point of interest, infinite curvature is only possible when the circle has zero radius (it reduces to a single point). In case of a $C^0$-connection between two parametric curves, it is obvious that the curvature goes to infinity. But apparently, this can also happen when two parametric curves meet with $C^1$-continuity. Any examples (e.g. using parametric polynomial curves)?

With regard to surfaces — the two principal curvatures $\kappa_1$ and $\kappa_2$ at the point $(u,v)$ of a parametric surface $X(u,v)$ are respectively the minimum and maximum normal curvature at this point $(u,v)$. The normal curvature in the direction $\phi$ is obtained when a plane containing the normal at $(u,v)$ is oriented in the direction $\phi$. The intersection of this plane and the surface results in a curve — the curvature of this curve in our point $(u,v)$ is called the normal curvature in the direction of $\phi$.

The Gaussian curvature $\kappa_G$ is simply the product of the two principal curvatures, the mean curvature $\kappa_M$ is the sum of the two principal curvatures.

If the Gaussian curvature is infinite, then at least one of the principal curvatures has to be infinite as well. I always thought that the curve (i.e. the intersection of the plane — oriented in the corresponding principal direction — and the surface) should therefore contain a kink, but apparently this does not have to be true. What am I missing?

Solution 1:

Short answer: the curvature is a property of the second derviative, and it's perfectly possible for a $C^1$ function to have a badly behaved second derivative.

In one variable, think about the function $f(t) = |t|^{3/2}$; it is $C^1$ but $f''(0) = +\infty$.

To turn this into a surface with infinite curvature, try $$X(u,v) = (u, |u|^{3/2}, v)$$ i.e. just extending this curve in the $z$ direction, forming a sheet. You'll find the normal curvature at the origin in the direction normal to the $xy$-plane is infinite.

Solution 2:

Consider the parametric curve given by $x(t) = t^3$, $y(t)=t^2$. Clearly this curve is $C_\infty$. However, at the point where $t=0$, there is a sharp corner in the curve, and its curvature is infinite (radius of curvature is zero).

The flaw in your thinking is related to parametrization. The derivatives of a curve depend on how it's parameterized, but radius of curvature is an intrinsic geometric property that is independent of parametrization. Therefore, you can't expect any hard-and-fast relationships between derivatives and radius of curvature. Everything breaks down at points where several derivatives are zero (which is what happens at $t=0$ on the example curve mentioned above).